S.chand publication solution of class 8 chapter 2 exercise 2 Exponents

EXERCISE 2


Q1 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper


Question 1

Evaluate:

(i) 2-3

Sol :

Using property an=1an

123=12×12×12

12×2×2=18


(ii) 60

Sol :

Using property: a0=1

60=1


(iii) (-3)-2

Sol :

Using property an=1an

1(3)2=13×13

13×3=19


(iv) 85 . 8-5

Sol :

Using property am×an=a(m+n)

⇒85 . 8-5 or

⇒85 × 8-5 

⇒8(5-5)

⇒80

⇒1



(v) 122

Sol :

Using property 1a=a1

⇒2(-2)×-1

⇒22

⇒4


(vi) 3(8-1)0

Sol :

Using property: a0=1

⇒3×1

⇒3


(vii) 121+131

Sol :

Using property : 1a1=a

⇒2+3

⇒5


(viii) 7x0

Sol :

Using property: a0=1

⇒7×1

⇒7


(ix) a2 ÷ a-2

Sol :

Using property : an=1an

a2÷1a2

a2×a21

⇒a2×a2

Using property : an×am=a(n+m)

⇒a4


(x) y6y4

Sol :

Using property : an=1a

1(y4)×(y6)

Using property: am×an=a(m+n)

1y(6+4)=1y(10) or

⇒y(-10)


(xi) (x -3)0

Sol :

Using property: (am)n=a(m×n)

⇒x0

⇒1


(xii) 60 + 6 -1 + 3 -1

Sol :

Using property: a0=1

⇒1+6 -1 + 3 -1

Using property: a1=1a

1+16+13

L.C.M of 6 and 3 is =2×3=6



26,333,31,1


1×6+1×1+1×26=6+1+26

or 96=32


(xiii) 1a0+b0

Sol :

Using property: a0=1

11+1=12


(xiv) (3x2)3

Sol :

Using property: (an)m=a(m×n)

⇒ 3(3)× x(2×3)

⇒ 27× x6

⇒ 27x6


(xv) [(57)6]0

Sol :

Using property: (am)n=a(m×n)

(57)6×0

(57)0

Using property : a0=1

⇒1


Q2 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 2

Simplify:

(i) x5 . x3

Sol :

Using property : am×an=a(m+n)

⇒x(5+3)

⇒x(8)


(ii) x8x3

Sol :

Using property : 1an=an

⇒x8 × x(-3)

Using property : am×an=a(m+n)

⇒x(8-3)

⇒x5


(iii) (x3)2

Sol :

Using property : am×an=a(m+n)

⇒x(3×2)

⇒x6


(iv) (3x2)3

Sol :

Using property : am×an=a(m+n)

⇒33×(x2)3

⇒27×x6

⇒27x6


(v) (5x-3yz2)-2

Sol :

Using property : an=1an

1(5x3yz2)2

152×x3×2×y2z2×2

125×x6×y2×z4

x625y2z4



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Question 3

Write the option that indicates the correct meaning of the given expression

(i) 4n3 means

(a) 4 times n + n + n

(b) 4n . 4n . 4n

(c) 4 . n . n . n

Sol :

⇒4n3

⇒4.n.n.n or (4×n×n×n)

Option (c) is correct


(ii) 6x3y2 means

(a) 6 . xy . xy . xy . xy . xy

(b) (6 . x. x. x . y)(7. x. x. x . y)

(c) 6 . x . x . x . y .y

Sol :

Option (c) is correct


Q4 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 4

Simplify:

(i) (a0b2)52a1

Sol :

(a0×5b2×5)2a1

(a0b10)2a1

Using property: 1a1=a

a0×a12b10

Using property: a0=1

a12b10


(ii) 4a3b02a2b1

Sol :

4a3b02a2b1

Using property: 1a=a1

⇒2a-3×b0× a-2 × b+1

Using property: an×am=a(m+n)

⇒2a-3-2×b0+1

⇒2a-5×b1

2ba5


(iii) 2x2y221x2y2

Sol :

Using property : 1a1=a1

⇒(2x2y-2)(21x-2y-2)

⇒(2×2)x2-2y-2-2

⇒4x0y-4

Using property : a0=1

⇒4y-4

4y4


(iv) (t4)3(t3)4

Sol :

Using property: (am)n=a(m×n)

t12t12

⇒1


(v) 4k2(4-1k+4k-2)

Sol :

⇒4k2(4-1k)+4k2(4k-2)

using property: 1a=a1

4k2×k4+4k2×4k2

k2+11+16k2k2

⇒k3+16


(vi) (3x22y1)2

Sol :

(3x2)2(2y1)2

Using property : (am)n=a(m×n)

32x4)22y2

91x4)41y2

4x4)9y2


(vii) [31(2)2]2

Sol :

(31)2{(2)2}2

Using property : (am)n=a(m×n)

32(2)4=916

 


Q5 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 5

Express each of the following as a power of 2 : 8 , 8x , 16x+3

Sol :

⇒8=2×2×2=23

⇒8x=(2×2×2)x=23x

⇒16x+3=(2×2×2×2)x+3


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Question 6

If a = 2m and b = 2{m+1} , show that 8a3b2=2m+1

Sol :

8a3b2

8(2m)3(2m+1)2

Using property : (am)n=a(m×n)

23×23m22m+2

Using property : a(m×n)=(am)n

23×22m+m22m+22

23×22m×2m22m×22

23×(22m)×2m(22m)×22

⇒2×2m

Using property : am×an=a(m+n)

⇒2m+1

Hence proved


Q7 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 7

If x = 2k and y = 2k+3 , what is the value of xy ?

Sol :

xy

2k2k+3

Using property: a(m+n)=am×an

(2k)(2k)+23

18


Q8 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 8

Write each expression such that there is no variable in the denominator .

(i) 3r2s312r3s7

Sol :

⇒(3r-2s3)(r3s-7)/12

Using property: am×an=a(m+n)

312(r2+3s37)

14(r1s4)

r1s44


(ii) 12x3y2z46x7y5z3

Sol :

126(x3y2z4)(x7y5z3)

Using property: am×an=a(m+n)

⇒2(x3-7×y-2+5×z4+3)

⇒2(x-4×y3×z7)

⇒2x-4y3z7


(iii) (5m)0n24mn3

Sol :

504(m0×n2)(m1n3)

Using property: a0=1

14(1×n2)(m1n3)

Using property: am×an=a(m+n)

14(n2+3)(m1)

nm14


(iv) 14a4k70a3k8

Sol :

1470(a4k)(a3k8)

Using property: a0=1

⇒14 (a-4k)(a-3k-8)

⇒14 (a-4-3k1-8)

⇒14 (a-7k-7)

 


Q9 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

Question 9

Simplify:

(i) 2n.82n1.162n43n

Sol :

2n×23(2n1)×24(2n)22(3n)

2n×26n3×28n26n

2n×(26n)×23×28n(26n)

⇒2-n× 2-3× 28n

Using property : am×an=a(m+n)

⇒2-n-3+8n

⇒2+7n-3


(ii) 2n+42.2n2.2n+3+23

Sol : 

Using property : am×an=a(m+n)

2n+42n+12n+4+23

(2n+4)(2n+4)2n+12n+4+23

12n+12(n+1)+(3)+23

1(2n+1)(2(n+1))×2(3)+23

1123+123

⇒1


(iii)  (0.6)0(0.1)1(323)1.(32)3+(13)1

Sol :

Using property: a0=1

1(110)1(323)1×(32)3+(13)1

Using property: a1=1a

1101(233)×(3323)(31)

110((8)(3))×((27)(8))3

993=96=32

 


Q10 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper



Question 10

If 9n×35×(27)33×(81)4=27 , then find the value of n

Sol :

9n×35×(27)33×(81)4

32n×35×3931×316

Using property: am×an=a(m+n)

32n+5+9316+1

32n+14317

⇒32n×314×3-17

⇒32n×314-17

⇒32n×3-3

32n33=27  according to question

⇒32n=33×33

⇒32n=36

⇒32n=32×3

⇒n=3


Q11 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper



Question 11

Show that :

(i) (xaxb)(a+b)×(xbxc)(b+c)×(xcxa)(c+a)=1

Sol :

Using property : aman=a(mn)

⇒{x(a-b)}(a+b) × {x(b-c)}(b+c) × {x(c-a)}(c+a)

Using property : (am)n=am×n

⇒x(a-b)×(a+b) × x(b-c)×(b+c) × x(c-a)×(c+a)

Using property : (a+b)(a-b)=a2-b2

xa2b2×xb2c2×xc2a2

Using property : am×an=a(m+n)

xa2b2+b2c2+c2a2

⇒x0

Using property : a0=1

⇒1

Hence proved


(ii) xa+b×xb+c×xc+a(xa×xb×xc)2=1

Sol :

Using property : am×an=a(m+n)

xa+b+b+c+c+a(xa+b+c)2

x2a+2b+2c(xa+b+c)2

(x2a+2b+2c)(x2a+2b+2c)

⇒1

Hence proved


Q12 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper



Question 12

Find x such that

(i) (74)3×(74)5=(74)3x2

Sol :

Using property : am×an=a(m+n)

(74)35=(74)3x2

(74)8=(74)3x2

On comparing both sides

⇒-8=3x-2

Transposing -2

⇒-8+2=3x

Transposing 3

x=63=2


(ii) (1258)×(1258)x=(52)18

Sol :

Using property : am×an=a(m+n)

(1258)x+1=(52)18

(5323)x+1=(52)18

(52)3(x+1)=(52)18

Comparing both sides

⇒3(x+1)=18

⇒3x+3=18

⇒3x=18-3

x=153=5

 


Q13 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper


Question 13

Find the value of x such that

(i) 32x1=127x3

Sol :

32x1=133(x3)

32x-1=3-3x+9

On comparing both sides

⇒2x-1=-3x+9

⇒3x+2x=9+1

⇒5x=10

x=105=2


(ii) (pq)3x+2=(qp)2x

Sol :

(pq)3x+2=(pq)2+x

On comparing both sides

⇒3x+2=-2+x

⇒3x-x=-2-2

⇒2x=-4

x=42=2


 

1 comment:

  1. please send mcqs and challege questions also

    ReplyDelete

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